Question: Find the slope and y-intercept of the line that is ${\text{perpendicular}}$ to $\enspace {y = -\dfrac{1}{2}x + 5}\enspace$ and passes through the point ${(-5, -8)}$. {1} {2} {3} {4} {5} {6} {7} {8} {9} {\llap{-}2} {\llap{-}3} {\llap{-}4} {\llap{-}5} {\llap{-}6} {\llap{-}7} {\llap{-}8} {\llap{-}9} {1} {2} {3} {4} {5} {6} {7} {8} {9} {\llap{-}2} {\llap{-}3} {\llap{-}4} {\llap{-}5} {\llap{-}6} {\llap{-}7} {\llap{-}8} {\llap{-}9}
Solution: Lines are considered perpendicular if their slopes are negative reciprocals of each other. The slope of the blue line is ${-\dfrac{1}{2}}$ , and its negative reciprocal is ${2}$ Thus, the equation of our perpendicular line will be of the form $\enspace {y = 2x + b}\enspace$ We can plug our point, $(-5, -8)$ , into this equation to solve for ${b}$ , the y-intercept. $-8 = {2}(-5) + {b}$ $-8 = -10 + {b}$ $-8 + 10 = {b} = 2$ The equation of the perpendicular line is $\enspace {y = 2x + 2}\enspace$. ${m = 2, \enspace b = 2}$